HMMT February, 2024

Solution:
We claim that if $n \geq 10$, then $2 n+3 \nmid 2^{n!}-1$ if and only if both $n+1$ and $2 n+3$ are prime.

If both $n+1$ and $2 n+3$ are prime, then assume $2 n+3 \mid 2^{n!}-1$. By Fermat's Little Theorem, $2 n+3 \mid 2^{2 n+2}+1$. However, since $n+1$ is prime, $\gcd(2 n+2, n!)=2$, so $2 n+3 \mid 2^{2}-1=3$, a contradiction.

If $2 n+3$ is composite, then $\varphi(2 n+3)$ is even and is at most $2 n$, so $\varphi(2 n+3) \mid n!$, done.

If $n+1$ is composite but $2 n+3$ is prime, then $2 n+2 \mid n!$, so $2 n+3 \mid 2^{n!}-1$.

The prime numbers between 50 and 100 are $53,59,61,67,71,73,79,83,89,97$. If one of these is $n+1$, then the only numbers that make $2 n+3$ prime are 53, 83, and 89, making $n$ one of 52, 82, and 88. These sum to 222.